Answer:
1) The polynomial in standard form is f(x) = x³ - 9·x² + 23·x - 15
2) The polynomial in standard form is f(x) = x³ - (2 - 2·i)·x² + 4·i·x
3) The polynomial in standard form is f(x) = x² + (3·i - 3)·x + 2 - 6·i
4) The polynomial in standard form is f(x) = x² - (5 + √5)·x + 6 + 3·√5
Explanation:
Given that that the polynomial is of least degree, we have;
The standard form is the form, f(x) = a·xⁿ + b·xⁿ⁻¹ +...+ c
1) The zeros of the polynomial are x = 5, 3, 1
Which gives the polynomial in factored form as_f(x) = (x - 5)·(x - 3)·(x - 1)
From which we have;
f(x) = (x - 5)·(x - 3)·(x - 1) = (x - 5)·(x² - 4·x + 3) = x³ - 4·x² + 3·x - 5·x² + 20·x - 15
f(x) = x³ - 9·x² + 23·x - 15
The polynomial in standard form is therefore f(x) = x³ - 9·x² + 23·x - 15
2) The zeros of the polynomial are x = 2, 0, 2·i
Which gives the polynomial in factored form as_f(x) = (x - 2)·(x)·(x - 2·i)
From which we have;
f(x) = (x - 2)·x·(x - 2·i) = (x² - 2·x)·(x - 2·i) = x³ - 2·i·x² + 2·x² + 4·ix
The polynomial in standard form is therefore f(x) = x³ - (2 - 2·i)·x² + 4·i·x
3) The zeros of the polynomial are x = 2, 1 - 3·i
Which gives the polynomial in factored form as_f(x) = (x - 2)·(x - 1 - 3·i)
From which we have;
f(x) = (x - 2)·(x - (1 - 3·i)) = x² - x + 3·i·x - 2·x + 2 -6·i = x² + 3·i·x - 3·x + 2 - 6·i
The polynomial in standard form is therefore f(x) = x² + (3·i - 3)·x + 2 - 6·i
4) The zeros of the polynomial are x = 3, 2 + √5
Which gives the polynomial in factored form as_f(x) = (x - 3)·(x - (2 + √5))
From which we have;
f(x) = (x - 3)·(x - (2 + √5)) = x² - 2·x - x·√5 - 3·x + 6 + 3·√5
f(x) = x² - 5·x - x·√5 + 6 + 3·√5 = x² - (5 + √5)·x + 6 + 3·√5
The polynomial in standard form is therefore f(x) = x² - (5 + √5)·x + 6 + 3·√5.